Binomial-Beta Claim Probability Components

Why Model Claim Probability Across Risks?

Suppose each insured has a fixed number of opportunities for a claim during a period. Even if each opportunity is a simple claim/no-claim trial, the risks need not have the same underlying probability of a claim. A useful model treats that probability as a risk-specific random quantity. Variation in outcomes for a given risk is then separated from variation in the underlying probabilities across risks.

This tutorial develops that separation for binomial outcomes with a Beta-distributed risk probability. The two quantities of interest are the expected process variance (EPV), which measures variation conditional on a risk’s probability, and the variance of hypothetical means (VHM), which measures how much the expected outcomes differ across risks. The results are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Let \(\Theta\) be a risk’s latent probability of a claim on each of \(m\) opportunities in a period. Assume \(\Theta\sim\operatorname{Beta}(\alpha,\beta)\), where \(\alpha>0\) and \(\beta>0\), and, conditional on \(\Theta=\theta\), the \(m\) claim indicators are independent Bernoulli trials with probability \(\theta\). The period claim count \(N\) then satisfies \(N\mid\Theta=\theta\sim\operatorname{Binomial}(m,\theta)\). The EPV for \(N\) is \(E[\operatorname{Var}(N\mid\Theta)]\); its VHM is \(\operatorname{Var}(E[N\mid\Theta])\).

The Beta distribution has density proportional to \(\theta^{\alpha-1}(1-\theta)^{\beta-1}\) for \(0<\theta<1\). Its parameters describe the distribution of risk probabilities across the population being modeled; they are not the number of trials for an individual. Write \(s=\alpha+\beta\). Then

E[\Theta]=\frac{\alpha}{s},\qquad \operatorname{Var}(\Theta)=\frac{\alpha\beta}{s^2(s+1)},\qquad E[\Theta(1-\Theta)]=\frac{\alpha\beta}{s(s+1)}.

The fixed number of opportunities \(m\) matters: the formulas below apply to a count over \(m\) trials with the same \(\Theta\). They do not directly apply if the number of trials is random, if trial probabilities vary within a risk, or if the risk probability changes during the period.

Deriving the Two Bühlmann Components

For a binomial count conditional on its probability, the mean is \(m\Theta\), and the variance is \(m\Theta(1-\Theta)\). Averaging the conditional variance over risks gives the EPV. Taking the variance of the conditional mean across risks gives the VHM.

\begin{aligned} \operatorname{EPV}(N) &=E[m\Theta(1-\Theta)] =m\frac{\alpha\beta}{s(s+1)},\\[4pt] \operatorname{VHM}(N) &=\operatorname{Var}(m\Theta) =m^2\frac{\alpha\beta}{s^2(s+1)}. \end{aligned}

Here is why the Beta moments used in these expressions hold. The Beta moment identity \(E[\Theta^r]=\alpha(\alpha+1)\cdots(\alpha+r-1)/(s(s+1)\cdots(s+r-1))\) gives \(E[\Theta]=\alpha/s\) and \(E[\Theta^2]=\alpha(\alpha+1)/(s(s+1))\). Therefore,

\begin{aligned} E[\Theta(1-\Theta)] &=E[\Theta]-E[\Theta^2]\\ &=\frac{\alpha}{s}-\frac{\alpha(\alpha+1)}{s(s+1)} =\frac{\alpha\beta}{s(s+1)},\\ \operatorname{Var}(\Theta) &=E[\Theta^2]-E[\Theta]^2 =\frac{\alpha\beta}{s^2(s+1)}. \end{aligned}

Multiplying the first expression by \(m\) averages the conditional binomial variance; multiplying the second by \(m^2\) accounts for the fact that the conditional mean is \(m\Theta\), not just \(\Theta\). This difference in exposure scaling is important when interpreting the components.

Theorem: By the law of total variance, the unconditional variance of the count is the sum of its EPV and VHM:

\operatorname{Var}(N) =E[\operatorname{Var}(N\mid\Theta)] +\operatorname{Var}(E[N\mid\Theta]) =\frac{m\alpha\beta(s+m)}{s^2(s+1)}.

The last form follows by adding the two components over a common denominator. The same decomposition applies to the claim rate \(R=N/m\), provided both components are scaled consistently: dividing the count by \(m\) divides each variance component by \(m^2\).

Quantity	Conditional mean	EPV	VHM
Count \(N\)	\(m\alpha/s\)	\(m\alpha\beta/[s(s+1)]\)	\(m^2\alpha\beta/[s^2(s+1)]\)
Rate \(R=N/m\)	\(\alpha/s\)	\(\alpha\beta/[m s(s+1)]\)	\(\alpha\beta/[s^2(s+1)]\)

The table’s rate row assumes the rate is the count divided by the fixed number \(m\) of trials. A count and a rate have different units, so their variance components should not be compared without accounting for that scaling. The VHM for the rate does not shrink as \(m\) grows: a risk’s underlying probability remains different from another risk’s even when each rate is measured with more trials.

Worked Example: Components for a Period Claim Count

A hypothetical book has risks with claim probabilities distributed as \(\Theta\sim\operatorname{Beta}(2,3)\). Each risk has \(m=10\) independent claim opportunities in a period, conditional on its probability. Find the mean count, EPV, VHM, and unconditional variance. No interest or mortality assumption is relevant to this frequency model.

Method. Use the binomial-Beta components for the count, keeping the conditional probability fixed across the 10 trials for each risk. Here \(\alpha=2\), \(\beta=3\), and \(s=5\).

The mean risk probability is \(E[\Theta]=2/5=0.4\). Thus the mean count is \(mE[\Theta]=10(0.4)=4\).

First calculate \(E[\Theta(1-\Theta)]=\alpha\beta/[s(s+1)]=(2)(3)/[(5)(6)]=6/30=0.2\). Therefore \(\operatorname{EPV}(N)=10(0.2)=2\).

Next, \(\operatorname{Var}(\Theta)=\alpha\beta/[s^2(s+1)]=6/[(25)(6)]=0.04\). Therefore \(\operatorname{VHM}(N)=10^2(0.04)=4\).

The unconditional variance is \(2+4=6\). As a check, the total-variance formula gives \(10(2)(3)(5+10)/[(5^2)(6)]=900/150=6\). The check agrees. A common mistake is to report only the binomial variance at the mean probability, \(10(0.4)(0.6)=2.4\); that omits the additional variance caused by differences in \(\Theta\) across risks.

Worked Example: Verifying the Total Variance

For another hypothetical group, let \(\Theta\sim\operatorname{Beta}(3,7)\), with \(m=10\) opportunities per risk. Calculate the EPV, VHM, and total variance, and check the result using the closed-form unconditional variance.

Method. This is a count model, not a model for the average rate. Set \(s=3+7=10\), with \(\alpha\beta=21\).

The expected process variance is \(\operatorname{EPV}(N)=10(21)/[(10)(11)]=210/110=21/11=1.9091\), to four decimal places.

The variance of the risk probability is \(\operatorname{Var}(\Theta)=21/[(10^2)(11)]=21/1100\). Hence \(\operatorname{VHM}(N)=10^2(21/1100)=2100/1100=21/11=1.9091\), to four decimal places.

Adding the components gives \(\operatorname{Var}(N)=42/11=3.8182\), to four decimal places. Independently, the closed form gives \(10(21)(10+10)/[(10^2)(11)]=4200/1100=42/11=3.8182\). The two calculations agree.

The mean probability is \(3/10=0.3\). Using only a binomial variance at that mean would give \(10(0.3)(0.7)=2.1\), not \(3.8182\). That calculation treats every risk as if its probability were exactly \(0.3\), removing the between-risk component.

Worked Example: Converting Count Components to Rate Components

In a hypothetical portfolio, \(\Theta\sim\operatorname{Beta}(2,2)\) and each risk has \(m=3\) claim opportunities. Calculate the count components and then the components for the observed frequency \(R=N/3\). Also verify the count variance using the three individual claim indicators.

Method. The Beta parameters give \(s=4\), \(E[\Theta]=2/4=0.5\), \(\operatorname{Var}(\Theta)=4/[(16)(5)]=1/20=0.05\), and \(E[\Theta(1-\Theta)]=4/[(4)(5)]=1/5=0.2\).

For the count, \(\operatorname{EPV}(N)=3(0.2)=0.6\), and \(\operatorname{VHM}(N)=3^2(0.05)=0.45\). Thus \(\operatorname{Var}(N)=0.6+0.45=1.05\).

Since \(R=N/3\), divide each count variance component by \(3^2=9\). This gives \(\operatorname{EPV}(R)=0.6/9=1/15=0.0667\) and \(\operatorname{VHM}(R)=0.45/9=0.05\), to four decimal places. The unconditional variance of the rate is \(1.05/9=7/60=0.1167\), to four decimal places.

For a separate check, each of the three claim indicators has unconditional variance \(0.5(1-0.5)=0.25\). Two distinct indicators have covariance \(\operatorname{Var}(\Theta)=0.05\), because they are independent conditional on \(\Theta\) but share that same latent probability. Therefore, \(\operatorname{Var}(N)=3(0.25)+2\binom{3}{2}(0.05)=0.75+0.30=1.05\), as before. The common error is to divide a rate component by \(m\) rather than \(m^2\); variances scale with the square of the multiplier.

Interpretation and Scope

The EPV describes uncertainty that remains even if the risk probability is known: a risk with a fixed probability can still experience a different number of claims from period to period. The VHM describes heterogeneity in expected counts among risks. A larger VHM means a larger spread of underlying probabilities under this model, while a larger EPV means more conditional outcome variation on average.

The Beta-binomial assumptions also explain why repeated outcomes for the same risk are not independent unconditionally. Conditional independence holds once \(\Theta\) is specified; averaging over a shared random \(\Theta\) produces positive dependence. In particular, two distinct claim indicators have covariance \(\operatorname{Var}(\Theta)\). This is one reason the unconditional variance can exceed the variance obtained by inserting the population mean probability into a single binomial variance formula.

When using the components, first identify the random variable being rated: a period count or a rate per opportunity. Then identify the number of trials and whether the same risk probability applies throughout. The formulas are not interchangeable with formulas for severity or aggregate loss, and they should not be used for varying trial probabilities without an appropriate model.

Key takeaway: With a fixed \(m\)-trial binomial count and \(\Theta\sim\operatorname{Beta}(\alpha,\beta)\), the EPV is \(m\alpha\beta/[s(s+1)]\), the VHM is \(m^2\alpha\beta/[s^2(s+1)]\), and their sum is the unconditional count variance. For a rate \(N/m\), divide both variance components by \(m^2\).

Check Your Understanding

For each problem, treat the stated Beta probability as shared across a risk’s trials, with conditional independence. Round decimal results to four places where needed.

A hypothetical risk probability has distribution \(\operatorname{Beta}(1,3)\), and each risk has \(m=4\) claim opportunities. Find the mean count, EPV, VHM, and unconditional variance. (Answer: mean \(=1.0000\), EPV \(=0.6000\), VHM \(=0.6000\), variance \(=1.2000\).)
For \(\Theta\sim\operatorname{Beta}(2,5)\) and \(m=7\), calculate the count EPV and VHM, then verify the total variance by adding them. (Answer: EPV \(=1.2500\), VHM \(=1.2500\), total variance \(=2.5000\).)
A portfolio has \(\Theta\sim\operatorname{Beta}(3,3)\) and six opportunities per risk. Calculate the mean count and both count variance components. (Answer: mean \(=3.0000\), EPV \(=9/7=1.2857\), VHM \(=9/7=1.2857\), total variance \(=18/7=2.5714\).)
Suppose \(\Theta\sim\operatorname{Beta}(1,4)\) and each risk has five opportunities. Find the EPV and VHM for the frequency \(R=N/5\), not for the count. (Answer: rate EPV \(=2/75=0.0267\), rate VHM \(=2/75=0.0267\), rate variance \(=4/75=0.0533\).)
A risk has \(\Theta\sim\operatorname{Beta}(2,3)\) and two claim opportunities. Verify the count variance by using the variances of the two indicators and their covariance. (Answer: each indicator has variance \(0.2400\), their covariance is \(0.0400\), and \(\operatorname{Var}(N)=2(0.24)+2(0.04)=0.5600\); equivalently EPV \(=0.4000\) plus VHM \(=0.1600\).)

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